Comparing Maclaurin Series of cos x and sin x

In summary, the conversation discusses differentiating and integrating the Maclaurin series versions of cos x and sin x. The general and specific terms for both functions are shown, and there is a discussion about whether they are correct. The conversation also includes a demonstration on how to simplify the series for sin x to equal cos x, and how to find the integral of cos x. A clarification is made on the notation for the integral of cos x, and a helpful tip is provided for typing math graphics.
  • #1
jackson_sun
11
0
I have been asked to differentiate cos x and six...the maclaurin series versions...

I have done the general and specific terms as shown below.

Im not sure if these are correct?
thanks

General Terms
cos x = ∑ (-1)n (x^(2n) / (2n)!)

COS x = ∑ (-1)n (x^(2n+1)/ (2n +1) / (2n)!)


sin x = ∑ (-1)n (x^(2n+1) / (2n+1)!)

‘sin x = ∑ (-1)n (((2n+1) x^(2n))/ (2n+1)!)

Specific Terms
sin x = x – (x^3 / 3!) + (x^5 / 5!) – (x^7 / 7!) + (x^9 / 9!) - …
‘sin x = 1 – (3x^2 / 3!) + (5x^4 / 5!) - (7x^6 / 7!) + (9x^8 / 9!) - …

cos x = 1 – (x^2 / 2!) + (x^4 / 4!) – (x^6 / 6!) + (x^8 / 8!) - …
COS x = x – (x^3 / 3 / 2!) + (x^5 / 5 / 4!) – (x^7 / 7 / 6!) + (x^9 / 9 / 8!) - …
thanks
 
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  • #2
You already posted the same thing else where, learn what the factorial actually means! Then you can simplify yours derivatives and realize what your not realizing right now!
 
  • #3
jackson_sun said:
I have been asked to differentiate cos x and six...the maclaurin series versions...

I have done the general and specific terms as shown below.

Im not sure if these are correct?
thanks

General Terms
cos x = ∑ (-1)n (x^(2n) / (2n)!)

COS x = ∑ (-1)n (x^(2n+1)/ (2n +1) / (2n)!)

You know to use "^" to indicate exponent of the "x"- use for the exponent on (-1) also: (-1)^n. Obviously "cos x" can't be equal to both of these. Do you mean cos' x? Also A/B/C, at best, ambiguous. Do you mean
((x^(2n+1)/(2n+1))/(2n)! or (x^(2n+1))/((2n+1)/(2n)!)?

Finally, if you are differentiating as you say, you are "going the wrong way"! The derivative of x^n is nx^(n-1), not (1/(n+1)) x^(n+1).


sin x = ∑ (-1)n (x^(2n+1) / (2n+1)!)

‘sin x = ∑ (-1)n (((2n+1) x^(2n))/ (2n+1)!)

Specific Terms
sin x = x – (x^3 / 3!) + (x^5 / 5!) – (x^7 / 7!) + (x^9 / 9!) - …
‘sin x = 1 – (3x^2 / 3!) + (5x^4 / 5!) - (7x^6 / 7!) + (9x^8 / 9!) - …

cos x = 1 – (x^2 / 2!) + (x^4 / 4!) – (x^6 / 6!) + (x^8 / 8!) - …
COS x = x – (x^3 / 3 / 2!) + (x^5 / 5 / 4!) – (x^7 / 7 / 6!) + (x^9 / 9 / 8!) - …
thanks
 
  • #4
It looks to me the cos x in capitals is the integral of cos x in normal...
 
  • #5
yeh COS x is the integral of cos x...i meant integral of cos x and derivative of sin x...sorry for the confusion.

I have realized how to simplify 'sin x so that it equals cos x...but i am not sure how to simplify the integral of cos x to equal sin x.

COS x = x – (x^3 / 3 / 2!) + (x^5 / 5 / 4!) – (x^7 / 7 / 6!) + (x^9 / 9 / 8!)
= ?

any idea? and on the subject...what is the notation for the integral of cos x...I have used COS x but this is obviously incorrect.
thanks...
 
  • #6
Take your second term...if you don't see the arithmetic, don't try this level of math yet, [tex]\frac{\frac{x^3}{3}}{2!} =\frac{x^3}{3\cdot2!} = \frac{x^3}{3!}[/tex].

I didn't take into account the sign of the integral, but you get what i mean.

To learn how to type the math graphics I am typing, click on the images I make appear. That will tell you what i typed to get that. Or at least say, "Integral of.."...
 
  • #7
I understand exactly what you have done.
How would it be done for the general terms is my bigger problem
 
  • #8
jackson_sun said:
I understand exactly what you have done.

Good
How would it be done for the general terms is my bigger problem

[tex]\cos x = \sum_{n=0}^{\infty} -1^n \frac{x^{2n}}{2n!}[/tex]

Integrate every term in the sum, with respect to x. In every individual term, n is constant.

So

[tex]\int\cos x dx= \sum_{n=0}^{\infty} -1^n \frac{\frac{x^{2n+1}}{2n+1}}{2n!}[/tex]

(2n+1)2n!=(2n+1)!,

therefore it simplifies to


[tex]\int\cos x dx= \sum_{n=0}^{\infty} -1^n \frac{x^{2n+1}}{(2n+1)!}[/tex], which is exactly your series for sin x.
 
  • #9
Just one comment: it would be far better to write
[tex]\int\cos x dx= \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}[/tex]
with parentheses around the -1. I would be inclined to interpret -1n as -(1n) which, of course, would be just -1!
 
  • #10
O yes i guess i would too, if someone else wrote that..thanks
 
  • #11
Thanks gents
 
  • #12
No Problem, and welcome to PF
 

1. What is a Maclaurin series?

A Maclaurin series is a type of power series expansion that represents a function as an infinite sum of terms involving powers of the independent variable. It is named after mathematician Colin Maclaurin, who discovered this method of expanding functions.

2. How are the Maclaurin series of cos x and sin x different?

The Maclaurin series of cos x and sin x are different in terms of the coefficients and the alternating pattern of signs. The series for cos x starts with positive coefficients, while the series for sin x starts with alternating positive and negative coefficients.

3. What is the formula for the Maclaurin series of cos x and sin x?

The Maclaurin series for cos x is 1 - x^2/2! + x^4/4! - x^6/6! + ..., while the Maclaurin series for sin x is x - x^3/3! + x^5/5! - x^7/7! + ....

4. How are the Maclaurin series for cos x and sin x related?

The Maclaurin series for cos x and sin x are related through the fundamental trigonometric identity cos^2 x + sin^2 x = 1. This relationship allows us to derive the series for one function from the series of the other.

5. What is the significance of comparing Maclaurin series of cos x and sin x?

Comparing the Maclaurin series of cos x and sin x allows us to understand the behavior and properties of these trigonometric functions. It also helps us approximate these functions and their values for different inputs, which is useful in applications such as physics, engineering, and finance.

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